Question

find the extreme values of the function on the given interval.
$f(x)=e^{x}\sin(x)$ on $0,\pi$
enter the maximum value here, or enter none if there isnt one.

enter the minimum value here, or enter none if there isnt one.

Explanation:

Step1: Find the derivative of \(f(x)\)

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = e^{x}\) and \(v=\sin(x)\).
\(u^\prime=e^{x}\), \(v^\prime=\cos(x)\)
\(f^\prime(x)=e^{x}\sin(x)+e^{x}\cos(x)=e^{x}(\sin(x)+\cos(x))\)

Step2: Find the critical points

Set \(f^\prime(x) = 0\), since \(e^{x}>0\) for all \(x\), we solve \(\sin(x)+\cos(x)=0\)
\(\sin(x)=-\cos(x)\), \(\tan(x)= - 1\) on \([0,\pi]\), \(x=\frac{3\pi}{4}\)

Step3: Evaluate the function at critical points and endpoints

• At \(x = 0\): \(f(0)=e^{0}\sin(0)=0\)
• At \(x=\frac{3\pi}{4}\): \(f(\frac{3\pi}{4})=e^{\frac{3\pi}{4}}\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}\)
• At \(x=\pi\): \(f(\pi)=e^{\pi}\sin(\pi)=0\)

Answer:

Maximum value: \(\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}\)
Minimum value: \(0\)